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Find all functions $f:\mathbb{R}_{\geq0}\rightarrow \mathbb{R}_{\geq0}$ which satisfies that for $x,y\in\mathbb{R}_{\geq0}$, $$f(x)+f(y)=f(x+y+2f(xy))$$

I spent quite some time trying to solve it but didn't succeed. It is clear that it has a solution $f(x) = 0$ and $f(x) = \sqrt{x}$ but the problem is whether they are all. Interesting related problem is that if we restrict $f$ to be continuous, or analytic, what will happen.

I tried to : Using Hamel basis to show the solutions are infinite. Failed. Also I tried to reduce it to Cauchy's 1-4 equations but didn't succeed. In the corse of it, I found interesting works of Aczel, Erdos and even Putnum, but they are not directly related, I guess.

Any idea? By the way, this is not a homework. A math fun came up with the problem but couldn't solve.

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1. Main Claim

We write $\mathsf{Sol}$ for the set of all solutions of the functional equation.

Claim. If $f \in \mathsf{Sol}$ is injective, then $f(x) = \sqrt{x}$.

Proof. Assume that $f \in \mathsf{Sol}$ is injective. Applying the functional equation in two ways, we find that

\begin{align*} f(x) + f(y) + f(1) &= f(x) + f(y + 1 + 2f(y)) \\ &= f(x + y + 1 + 2f(y) + \boxed{ 2f(xy + x + 2xf(y)) }) \end{align*}

and

\begin{align*} f(x) + f(y) + f(1) &= f(x + y + 2f(xy)) + f(1) \\ &= f(x + y + 2f(xy) + 1 + 2f(x + y + 2f(xy))) \\ &= f(x + y + 1 + \boxed{ 2f(xy) + 2f(x) } + 2f(y)). \end{align*}

By the injectivity assumption, we have

\begin{align*} f(xy + x + 2xf(y)) &= f(xy) + f(x) \\ &= f(xy + x + 2f(x^2y)). \end{align*}

Stripping $f$ off both sides of the identity above, we find that

$$ f(x^2y) = xf(y). $$

So it follows that $f(x) = f(1)\sqrt{x}$, and plugging this back to the functional equation shows that $f(1) = 1$. Therefore $f(x) = \sqrt{x}$. ////


2. Some extra observations

Lemma 1. Let $f \in \mathsf{Sol}$. Then

  1. $f(x) + nf(1) = f(x + 2nf(x) + n + n(n-1)f(1))$ for all $x\geq0$ and $n \geq 0$.
  2. $f(0) = 0$.

Proof. (1) Let us write $x_n = x + 2nf(x) + n + n(n-1)f(1)$. Since $x_0 = x$, the $n = 0$ case is trivial. Now assume that $f(x) + nf(1) = f(x_n)$. Then

\begin{align*} f(x) + (n+1)f(1) &= f(x_n) + f(1) \\ &= f(x_n + 1 + 2f(x_n)) \\ &= f(x_n + 1 + 2f(x) + 2nf(1)) = f(x_{n+1}) \end{align*}

and hence the claim follows from the mathematical induction.

(2) From the previous part with $x = 0$ and $n = 2$, it follows that

$$ f(0) + 2f(1) = f(4f(0) + 2 + 2f(1)). $$

Now using the identity $f(x + 2f(0)) = f(0) + f(x)$ twice, we find that

$$ f(4f(0) + 2 + 2f(1)) = 2f(0) + f(2 + 2f(1)) = 2f(0) + 2f(1). $$

Comparing two formulas yields $f(0) = 0$. ////

Corollary 2. If $f \in \mathsf{Sol}$ is non-decreasing, then

  1. $\lim_{x\downarrow 0} f(x) = 0$.
  2. Either $f\equiv 0$ or $f(x) = \sqrt{x}$.

Proof. (1) Write $\ell = \lim_{x\downarrow 0}f(x)$. We claim that $\ell = 0$. Indeed, let $\epsilon > 0$ and write

\begin{align*} \epsilon_1 &= 2f((2+2f(1))\epsilon) - 2\ell + \epsilon, \\ \epsilon_2 &= 2f((2 + 2f(1) + 2\ell + \epsilon_1)\epsilon) - 2\ell + \epsilon_1 + \epsilon. \end{align*}

Then mimicking the previous proof, we have

\begin{align*} 2\ell + 2f(1) &\leq 2f(\epsilon) + f(2+2f(1)) \\ &= f(\epsilon) + f(2 + 2f(1) + 2\ell + \epsilon_1 ) \\ &= f(2 + 2f(1) + 4\ell + \epsilon_2) \\ &\leq f(2 + 2f(1) + 4f(\epsilon_2) + \epsilon_2) = f(\epsilon_2) + 2f(1). \end{align*}

Here, intermediate equalities follow from the functional equation together with the definition of $\epsilon_1$ and $\epsilon_2$, and the last equality follows from Lemma 1.(1). Since $\epsilon_2 \downarrow 0$ as $\epsilon \downarrow 0$, taking $\epsilon \downarrow$ to the above bound tells that $2\ell + 2f(1) \leq \ell + 2f(1)$ and hence $\ell = 0$.

(2) If $f(a) = 0$ for some $a > 0$, then $f(2a) \leq f(2a+2f(a^2)) = 2f(a) = 0$ and repeating this procedure tells that $f(2^n a) = 0$ for all $n \geq 0$. Since $f$ is non-decreasing, this implies that $f \equiv 0$.

So we may assume that $f(x) > 0$ for all $x > 0$. Then for any $0 \leq x < y$, there exists $t > 0$ such that $y - x > t + 2f(t x)$ by the previous part. Then

$$ f(y) \geq f(x + t + 2f(t x)) = f(x) + f(t) > f(x) $$

and hence $f$ is strictly increasing. Therefore by the main claim, $f(x) = \sqrt{x}$. ////

Corollary 3. If $f \in \mathsf{Sol}$ is continuous, then either $f\equiv 0$ or $f(x) = \sqrt{x}$.

Proof. In view of Corollary 2, it suffices to show that $f$ is non-decreasing. Indeed, fix $x \geq 0$. Since $t \mapsto t + 2f(xt)$ is continuous and takes every value in $\mathbb{R}_{\geq 0}$, for each $y > x$ we can choose $t > 0$ such that $t + 2f(xt) = y - x$. Then

$$ f(y) = f(x + t + 2f(xt)) = f(x) + f(t) \geq f(x). $$

Therefore $f$ is non-decreasing and hence the conclusion follows. ////

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  • 2
    $\begingroup$ This was very educative! Thanks $\endgroup$ – Hari Shankar Jan 25 '18 at 2:25
  • $\begingroup$ Very interesting. Still I feel the real answer (for all real functions) are infinite. Thank you very much! $\endgroup$ – Yoriyuki Yamagata Jan 29 '18 at 21:54
  • $\begingroup$ @YoriyukiYamagata Glad it helped! :) For the general case I have no affirmative answer, and in fact I feel ambivalent about which will be the case. One one hand, the restriction $f(x) \geq 0$ for $x \geq 0$ is strong enough to force any solution of the Cauchy functional equation to be linear and one might hope that this is also true in your problem. On the other hand, the equation seems less powerful than the Cauchy functional equation and might allow some exotic solutions. I would love to see if we can have an answer. $\endgroup$ – Sangchul Lee Jan 29 '18 at 22:27
  • $\begingroup$ Hmm... Let the function $f(x) = \sqrt{x}$ if $x$ is algebraic while $f(x) = 0$ if $x$ is transcendental? Is this another solution? $\endgroup$ – Yoriyuki Yamagata Jan 30 '18 at 5:15
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    $\begingroup$ @YoriyukiYamagata, Notice that if $\alpha \in (0, 1)$ is transcendental, your function gives $f(\alpha) + f(1-\alpha) = 0$ while $f(\alpha + (1-\alpha) + 2f(\alpha(1-\alpha))) = f(1) = 1$. My humble guess is that there is no non-trivial solution which is measurable. $\endgroup$ – Sangchul Lee Jan 30 '18 at 10:56

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